The Computational Universe: Debunking the Myth or Proving the Possibility?

[Coin]

Well, I think by computable here it means that you'd have to simulate it perfectly. I think sirchasm's objection was that this isn't ever done in reality.

But naturally it could be estimated by taking a hypothetical Turing machine and let it run to infinity: an incomputable system won't even finish as t->infinity, while a computable system will. So PI, for instance, is an incomputable quantity because there is no last digit. I think this is what you meant?

Um, you may or may not have some names flipped here?

Anyway, I could buy the notion of an algorithm which "simulates" a dynamical system by producing successively better approximate simulations of the system as it runs forever. And this would be consistent with the notion of "computable" as in "computable number", I.E., a computable number is one (like PI) that a TM can produce in the limit as its runtime goes to infinity.

I'm not sure though whether or not this notion of "computing" is totally consistent with sirchasm's original question, which had something to do with whether the universe "is a computer"...

 

[sirchasm]

It's not a result, though. It's many separated results due to local interactions that can have no input from causally-disconnected regions.

The observable fact that regions of the current universe are disconnected informationally isn't an outcome for a single expanding universe, you mean?

how can expansion be considered anything remotely like computation?

Well, we would need to build a logical machine that can prove: "expansion is not computable, therefore cannot be a computation", I guess.

 

[sirchasm]

Although entanglement with entropy looks like at most a connection. a distance operator that looks fixed; it does encode something, which is a direction or an angle in 'entanglement space'. The dimension of the connection is 1, but the signal is "there is a connection", this signal has a different limit (it must have, since it's FTL, effectively). The distance operator also encodes a rotational symmetry or SU(2). ??

The "yes, there is a connection' signal is probabilistic, we can set this to 1 locally, but between two 'near' states.
Extending this to 'distant' states is a bit of a problem (nyar ...nyar)

 

[sirchasm]

Well we've thrashed the idea of building a machine and physicality; done the entanglement thing; and still here.

So can we build a 'circuit' with entanglement? Sure, we already do.
Can we measure distant entanglement? we have to prepare something - we have to make a local pair of massive (or momental) quanta 'coherent' -i.e. aligned in the independent-of-distance entangled space.

But there is the polarization in the CMB; if we can build a device = a circuit, that measures this, we have a 'fix' on the content of an entropy-space.

If we consider the angle that entanglement encodes as a polarizable potential, then we need a machine that can 'expand' and 'deform' it - by re-polarizing it.

 

[Chalnoth]

Well, entanglement is just about maintaining particles in a quantum superposition of states. This is important for quantum computers because any time you collapse a superposition, you lose information about the other components of the superposition.

It isn't about information transfer between separate parts of the quantum computer (you can't do that with entanglement). It's about preventing information loss due to interactions.

 

[sirchasm]

See how I'm using "potential" abstractly, it's just "something we might use" in this context.
Polarization of EM fields and radiation (from small to large wavelengths = distance operators on momentum), are connected to entanglement, since we can entangle photons.

There's a 'tanglement' potential which we entangle = connect; this is the machine needed. Entangled photons, then have a distance operator which acts in a 'moment-free' space, as a photon's distance acts in a 'mass-free' one. Down-converting is an operator that rotates a potential in photons (in their momentum) so it's now at right angles to itself as 2 photons, 1 e-bit. This implies a photon is entangled 'with' itself; a photon implies a rotational group on U(1) called SU(2), because mass-charge is scalar for all d.o.f. in wherever e-space is (obviously it's a subspace of momentum-space).

We can derive entangled states from momentum (of mass charge spin) as exchanges along mass-spin and charge-spin directions (pair-production and Hawking-Bekenstein radiation, q-bits or 'qubits'); spin-spin directions (Stern-Gerlach, neutron interferometry, QH-effect, AB effect, q-bits); but not mass-charge, because [h, \hbar]\; has to encode entanglement; mass-charge terms are scalar in e-space.

QIS calls an entangled quantum 'direction' an e-bit, which interacts with q-bit representations (atoms, electrons, photons); we haven't managed to encode e-bits in terms of quasiparticles yet. Or have we?

 

[Chalnoth]

See how I'm using "potential" abstractly, it's just "something we might use" in this context.
Polarization of EM fields and radiation (from small to large wavelengths = distance operators on momentum), are connected to entanglement, since we can entangle photons.

There's a 'tanglement' potential which we entangle = connect; this is the machine needed. Entangled photons, then have a distance operator which acts in a 'moment-free' space, as a photon's distance acts in a 'mass-free' one. Down-converting is an operator that rotates a potential in photons (in their momentum) so it's now at right angles to itself as 2 photons, 1 e-bit. This implies a photon is entangled 'with' itself; a photon implies a rotational group on U(1) called SU(2), because mass-charge is scalar for all d.o.f. in wherever e-space is (obviously it's a subspace of momentum-space).

We can derive entangled states from momentum (of mass charge spin) as exchanges along mass-spin and charge-spin directions (pair-production and Hawking-Bekenstein radiation, q-bits or 'qubits'); spin-spin directions (Stern-Gerlach, neutron interferometry, QH-effect, AB effect, q-bits); but not mass-charge, because [h, \hbar]\; has to encode entanglement; mass-charge terms are scalar in e-space.

QIS calls an entangled quantum 'direction' an e-bit, which interacts with q-bit representations (atoms, electrons, photons); we haven't managed to encode e-bits in terms of quasiparticles yet. Or have we?

Honestly, I can't tell whether you're just writing gibberish or you merely haven't explained yourself properly.

 

[sirchasm]

Well, you see, I'm 'talking' in code.
What would you describe the potential in EM radiation as? The gauge of QFT?
Can you describe the connection or disconnection between nonlocality, superposition and measurement?

 

[Chalnoth]

Well, you see, I'm 'talking' in code.

Uh-huh.

What would you describe the potential in EM radiation as?

The potential is a function which, when differentiated, gives the value of the EM field.

The gauge of QFT?

A gauge transformation is a way in which a function in the theory can be varied without changing the physical effects of the theory. For example, since the physical stuff in EM is the field, and since the derivative of the potential is the field, you can add to the potential any function that has a derivative of zero. Different choices of the function added to the potential are different choices in gauge.

The way the word gauge is used throughout QFT, as well as gravity, is essentially the same as this. In General Relativity, for instance, a coordinate transformation is also a gauge transformation: whether you look at a system with one set of coordinates or another, the physics is still the same. So a change in coordinates doesn't change the physics, and is sometimes called a gauge transformation.

In QFT, the gauge is very important when, for example, you do a path integral. With a path integral, you're integrating over different paths. But some of those paths are going to be related to one another by a gauge transformation: they're actually the same path. So you have to take the gauge invariance of the theory into account when performing certain calculations if you want to be sure to not double-count things.

Can you describe the connection or disconnection between nonlocality, superposition and measurement?

There is no non-locality in quantum physics, as quantum physics is a fully local theory.

 

[sirchasm]

There is no non-locality in quantum physics, as quantum physics is a fully local theory.

Do you have references to articles that confirm the non-nonlocality of quantum transportation of states? Or that confirms Bell's experiments are entirely local, or the results of any quantum experiment that can determine the local states of double-slit experiments, at all times?

The latest is: QM is a nonlocal theory of quantum states; nonlocal because time and space are in an independent frame - our frame.

A gauge transformation ...

What does a gauge transformation transform? I asked what the gauge is, not what transforming a gauge is. The photon is the gauge of the EM field; what is "the photon"?

So, what is THE connection between superposition, nonlocality and measurement? Three things, one connection? Or isn't there one?

 

[Chalnoth]

Do you have references to articles that confirm the non-nonlocality of quantum transportation of states? Or that confirms Bell's experiments are entirely local, or the results of any quantum experiment that can determine the local states of double-slit experiments, at all times?

Meh, I don't see why I should. Intrinsic non-locality is patently absurd. But regardless, this is off-topic for the thread and for the forum.

What does a gauge transformation transform? I asked what the gauge is, not what transforming a gauge is.

A gauge is a type of transformation.

The photon is the gauge of the EM field; what is "the photon"?

Hardly. When talking about a "gauge boson", the word "gauge" is an adjective to describe the boson. It is not a noun. The reason it is used is because the interactions between gauge bosons and matter can be derived just by stating that they each transform under the right sort of gauge symmetry.

 

[sirchasm]

Intrinsic non-locality is patently absurd

Then, it shouldn't be difficult to explain why locality is patently non-absurd?

Quantum electrodynamics is an abelian gauge theory with the symmetry group U(1) and one gauge field, the electromagnetic field, with the photon being the gauge boson.

Why doesn't this say "the photon being the boson gauged by the field, it's a gauged boson"?

You mean the photon is a gauge of the field = it's a transformation of the e and u in the field?

 

[Chalnoth]

Then, it shouldn't be difficult to explain why locality is patently non-absurd?

Because locality doesn't involve faster-than-light transmission of information. Which entanglement in QM doesn't do, by the way.

Why doesn't this say "the photon being the boson gauged by the field, it's a gauged boson"?

Why would it?

You mean the photon is a gauge of the field = it's a transformation of the e and u in the field?

No. The photon is a quantum of the field. "Gauge of the field" doesn't make any sense.

 

[sirchasm]

locality doesn't involve faster-than-light transmission of information.

(and, this is why locality isn't absurd?

A gauge is a type of transformation. ... The photon is a quantum of the field. "Gauge of the field" doesn't make any sense.

But a photon is "the gauge of the EM field". How many published articles would you estimate I can locate that say exactly those 6 words? Are these papers nonsensical??
You say: it's a transformation, which implies it's 'gauged', not 'a gauge'...

Nonlocality in quantum mechanics

[edit] Einstein, Podolsky and Rosen

Main article: EPR Paradox

In 1935, Einstein, Podolsky and Rosen published a thought experiment [2] with which they hoped to expose the inadequacies of the Copenhagen interpretation of quantum mechanics in relation to the lack of determinism at the microscopic scale that it described.
In particular, they hoped to demonstrate that the probabilistic nature of the results of measurements on particles could be described through the means of some ‘hidden’ variables that predetermine the result of a measurement, but to which an observer does not have access.

In physical terms, this experiment can be represented as a spin-zero particle decaying into two spin-half particles such that there is no interaction between the two particles after decay. Since spin is a conserved quantity, measurements of spin on the two particles must anti-correlate. The quantum state of the two particles prior to measurement can be written as[3]

\left|\psi_{AB}\right\rangle=\frac{1}{\sqrt{2}} \bigg(\left|\uparrow\right\rangle_A \left|\downarrow\right\rangle_B + \left|\downarrow\right\rangle_A \left|\uparrow\right\rangle_B \bigg)

Here, subscripts A and B distinguish the two particles, though it is more convenient and usual to refer to these particles as being in the possession of two experimentalists called Alice and Bob.

This state allows us to predict the probability of a particular result. Alice, for example, will measure her particle to be spin-up in an average of fifty percent of measurements. However, when Alice measures her particle it causes the state to collapse so that if Alice measures spin-up, Bob must measure spin-down and vice versa. Hence, either party is capable of setting the spin of the other’s particle instantaneously. Such behaviour is non-local because the measurement of one particle is able to influence the physical state of another independently of the distance between them, so that no information could travel between them.

Einstein, Podolsky and Rosen saw this as evidence of a causal effect propagating at superluminal speeds, which is in violation of the laws of special relativity[2]. They further pointed out that such an unwanted result could be avoided by admitting the presence of hidden variables that determine the results of measurements for each of the entangled particles, which would restore locality to physics.)

BUT (that's a big but), there are no 'hidden variables'; this is nonsensical...

 

[Chalnoth]

(and, this is why locality isn't absurd?

Basically, yes. Non-locality, in essence, requires magic.

But a photon is "the gauge of the EM field". How many published articles would you estimate I can locate that say exactly those 6 words? Are these papers nonsensical??

Quite a few. But find me one that says it's a photon.

You say: it's a transformation, which implies it's 'gauged', not 'a gauge'...

A gauge transformation is a type of transformation. A gauge field is a field which is invariant under this sort of transformation. A gauge boson is a quantum of a field that is invariant under this sort of transformation. In all situations, the word 'gauge' is an adjective.

BUT (that's a big but), there are no 'hidden variables'; this is nonsensical...

Locality is restored if you just drop the postulate of wavefunction collapse from quantum mechanics. It's not needed.

 

[sirchasm]

Non-locality, in essence, requires magic.

Then EPR, Stern-Gerlach, double-slit and any experiments with nonlocal effects demonstrate magic.

When you say "locality is restored", do you mean you know where each electron is that passes through a double-slit, or any interferometer?

And if a photon is a gauge transformation, is it scalar? The gauge is a scalar quantity?
I don't see how I can transform an adjective into anything except another adjective, unless "transformation" of the adjective makes it something else.

"Hot" is an adjective = a gauge ; I can do a "hot" transformation, into something else? What else, it doesn't even make sense??

 

[sirchasm]

If I use "hot", I can differentiate objects - the transformation - into more and less "hot". It's a gauge that can make sense.
If I want to transform an area into a known quantity, I might measure or calculate it with angles and distances.

The gauge I might use could be sticks stuck in the ground, or say rocks. I could use both, the gauge is invariant because it "transforms" the area into a bounded one, whether I use sticks or rocks or something else as a marker (of position).

 

[Chalnoth]

This is too far off topic for me. I'm done.

 

[sirchasm]

Well, that could mean I've either presented an argument you can't answer; or it could mean I am 'off-topic'.

If the latter, my pick is I've got to here, because you asked me to, in order to argue your position.
So if you withdraw your argument (after posting it), I'll get back to the topic (I'll have another look at the OP, and see if I can remember why I posted it)

 

[sirchasm]

Note: gauge theories treat fundamental objects, like the photon. There are gauges for all kinds of things - after all it's just a way to measure something.

However the theories use gauges that represent fundamental physical measurements - which are invariant for some aspect of the space which is gauged. The photon is an invariant gauge for EM theory in a quantum basis (since it's a quantized one).

 

[cristo]

This is too far off topic for me. I'm done.

I agree. This thread's done.